ok i changed my opininon. could you help m with only one problem. i havent solve it. that is in the chapter 3.additional problem 28 ;
i would try to write a summary of the problem : assume an ice cube retains its cubical shape as it melts. if we call edge lenth s its vlume is v=s^3 and the surface area is = 6*s^2. we also assume that the cube's volume decreases at a rate that is proportional to its surface area. in math terms : dv/dt=-k(6*s^2) ; assume that the cube lost 1/4 of its volume during the first hour and that the volume is Vo at t=0. how long will it take the ice cube to melt...

i tried to solve that with integrating(as t goes to 0 V goes to 3V/4) and tried to find k but then k is going to be a strange value.also the chapter is about derivative applications i can not solve it with integral. then what should i do please help me with this problem...

Okay, first of all:
Here, it is smart to express the surface S in terms of the volume V:
[tex]S=6V^{\frac{2}{3}}[/tex]
Thus, the differential equation for the rate of change of the volume is:
[tex]\frac{dV}{dt}=-6kV^{\frac{2}{3}}[/tex]
This is a separable equation:
[tex]\frac{dV}{V^{\frac{2}{3}}}=-6kdt[/tex]
or, integrating both sides from t=0 and and t=T:
[tex]3(V(T)^{\frac{1}{3}}-V(0)^{\frac{1}{3}})=-6kT[/tex]
or simply, for arbitrary T:
[tex]V(T)=(V(0)^{\frac{1}{3}}-2kT)^{3}[/tex]
Now you should be able to do the last steps on your own!

yes i know first put V(t)=3v/4 t=1 find k than put v(t)=0 put k and find t. this what i must do isn't it?. but this is a problem from derivative chapter. but we found the answer with integration is it true?